Question

evluate i^-50 where I is iota


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Answer 1

Correct question:

Evaluate i⁻⁵⁰ where i is a complex number or imaginary number

Solution:

As we know that

i is a complex number which is equal to √-1

Now , i⁻⁵⁰ can be written as

→ i⁻⁵⁰

We know that

a⁻ⁿ = 1/aⁿ

Similarly

→ 1/i⁵⁰

→ 1/(i⁴⁸ × i²)

Here , i is raised to any multiple of 4 is equal to 1 & i = √-1

→ 1/1 × (√-1)²

→ 1/1 × (-1)

→ 1/(-1)

→ - 1

Hence , i⁻⁵⁰ = - 1

Answer 2

$\large \bf \red{ \underline{ \underline{To\:Find:-}}}$

• $\sf the \: value \: of \: {(i)}^{ - 50}$

$\large \bf \green{ \underline{ \underline{Solution:-}}}$

As we know that :-

The value of i (iota) = √-1

Squaring on both sides

$\sf \implies {i}^{2} = { (\sqrt{ - 1}) }^{2}$

Now we have to find the value of $\sf {(i)}^{ - 50}$

By the identity

$\sf \rightarrow {a}^{ - n} = \dfrac{1}{ {a}^{n} }$

$\sf \rightarrow {i}^{ - 50} = \dfrac{1}{ {i}^{50} }$

$\sf \rightarrow \dfrac{1}{ {i}^{50} } = \dfrac{ { 1} }{ { {(i}^{2} )}^{25} }$

As i² = -1

$\sf \rightarrow \dfrac{1}{ { {(i}^{2}) }^{25} } = { (- 1)}^{25}$

We know that -1^even integers = 1 and -1^odd integers = -1

As 25 is an odd number

$\sf \rightarrow { (- 1)}^{25} = - 1$

$\large \boxed{ \sf \purple{\rightarrow { (i)}^{ - 50} = - 1}}$